Question

If y=log root over 1+tanx/1-tanx, prove that dy/dx =sec2x

Answer 1

$$\begin{gathered} y=\log \sqrt{\frac{1+\tan x}{1-\tan x}} \\ Sothiscanbewrittenasy=\frac{1}{2}\left\{\log \right(1+\tan x)-\log (1-\tan x\left)\right\} \\ Sodifferntiatingit,weget\frac{dy}{dx}=\frac{1}{2}\left\{\frac{d}{dx}\right(\log (1+\tan x))-\frac{d}{dx}(\log (1-\tan x)\} \\ =\frac{1}{2}\{\frac{1}{1+\tan x}\times se{c}^{2}x-\frac{1}{1-\tan x}\times (-se{c}^{2}x)\} \\ =\frac{1}{2}\{\frac{1}{1+\tan x}\times se{c}^{2}x+\frac{1}{1-\tan x}\times \left(se{c}^{2}x\right)\} \\ =\frac{se{c}^{2}x}{2}\{\frac{1}{1+\tan x}+\frac{1}{1-\tan x}\} \\ =\frac{se{c}^{2}x}{2}(\frac{1-\tan x+1+\tan x}{1-{\tan }^{2}x}) \\ =\frac{se{c}^{2}x}{2}(\frac{2}{1-{\tan }^{2}x})=\frac{se{c}^{2}x}{1-{\tan }^{2}x}=\frac{1+{\tan }^{2}x}{1-{\tan }^{2}x}=\frac{1}{\cos 2x}=sec2x \end{gathered}$$